# creating delimiters and inequality signs

I would like to have a delimiter whose upper half is the upper half of the delimiter [ and whose lower half is the lower half of the delimiter ( -- so that it be one continuous line, and also the matching delimiter whose upper half is the upper half of the delimiter ) and whose lower half is the lower half of the delimiter ] .

Correspondingly, I would like to have a binary (inequality) relation with the full-size sign > over the succession of a half-length = sign followed on the right by an empty space of the same horizontal extent as the half-length = sign, so that the the total horizontal extent of the combination of the the half-length = sign plus the empty space be the same as that of the full-size sign > . Such a binary relation could be called "greater than or half-equal to". Similarly, I would like to have a "less than or half-equal to" sign, which is a binary relation with the full-size sign < over a half-length = sign preceded on the left by an empty space of the same horizontal extent as the half-length = sign.

Can this be done without METAFONT (which I don't know at all)? Thank you very much for your help.

• Should this be an extensible delimiter, or just a regular-sized one? – Werner Aug 26 '13 at 2:59
• have these symbols appeared somewhere in a published work? if so, can you give a specific reference? with good documentation, i can submit them to unicode for consideration, and likely acceptance, which flows ultimately into actual fonts. – barbara beeton Aug 26 '13 at 22:17
• Werner asks: "Should this be an extensible delimiter, or just a regular-sized one?" -- Extensible would be better, but a regular one would do for now. – Iosif Pinelis Aug 27 '13 at 4:02
• Barbara Beeton asks: "Have these symbols appeared somewhere in a published work?" -- No, not yet, but they should be used, I think. – Iosif Pinelis Aug 27 '13 at 4:04

The “asymmetric” square brackets are \lfloor, \rfloor, \lceil and \rceil.

For the strange inequalities, here is some code:

\documentclass{article}
\usepackage{trimclip}

% the left half of =
\newcommand{\lhalfeq}[1]{\clipbox{0pt 0pt {.5\width} 0pt}{$#1=$}}
% the right half of =
\newcommand{\rhalfeq}[1]{\clipbox{{.5\width} 0pt 0pt 0pt}{$#1=$}}
% greater than or half-equal to
\newcommand{\gtheq}{\mathrel{\mathpalette\xgtheq\relax}}
\newcommand{\xgtheq}[2]{%
\vcenter{\hbox{%
\oalign{$#1>$\cr\lhalfeq{#1}\hidewidth\cr}%
}}
}
% less than or half-equal to
\newcommand{\ltheq}{\mathrel{\mathpalette\xltheq\relax}}
\newcommand{\xltheq}[2]{%
\vcenter{\hbox{%
\oalign{$#1<$\cr\hidewidth\rhalfeq{#1}\cr}%
}}
}

\begin{document}
$a\gtheq b\ltheq c$
\end{document}


Thanks to Heiko, a slightly better implementation for vertically centering the symbol.

\documentclass{article}
\usepackage{trimclip}

% the left half of =
\makeatletter
% the right half of =

\sbox0{$#1\vcenter{}$}%
\raisebox{\dimexpr\height-2\ht0\relax}[2\dimexpr\height-\ht0\relax][0pt]{$\m@th$=}%
}
% greater than or half-equal to
\newcommand{\gtheq}{\mathrel{\mathpalette\xgtheq\relax}}
\newcommand{\xgtheq}[2]{%
\vcenter{\hbox{%
\oalign{$\m@th#1>$\cr\lhalfeq{#1}\hidewidth\cr}%
}}
}
% less than or half-equal to
\newcommand{\ltheq}{\mathrel{\mathpalette\xltheq\relax}}
\newcommand{\xltheq}[2]{%
\vcenter{\hbox{%
\oalign{$\m@th#1<$\cr\hidewidth\rhalfeq{#1}\cr}%
}}
}
\makeatother

\begin{document}
$a\gtheq b\ltheq c$
\end{document}


• The symbol is not centered, because it also contains the space from the equal sign to the base line at the bottom of the symbol. It can be fixed by replacing $#1=$ with \sbox0{$#1\vcenter{}$}\raisebox{\dimexpr\height-2\ht0\relax}[2\dimexpr\height-\ht0\relax][0pt]{$\m@th=$}. – Heiko Oberdiek Aug 26 '13 at 22:32
• The answer by egreg and the comment by Heiko were very helpful to me. Yet, I still have some problems, as described in my answer below. – Iosif Pinelis Aug 27 '13 at 4:08

Is this what you asked for? :

$\left\lfloor abc \right\rfloor \left\lceil xyz \right\rceil$


• Ideally, I would like a combination of a bracket and a parenthesis, as I described. – Iosif Pinelis Aug 27 '13 at 4:09

Thank you for your answers; egreg's answer was especially useful. I was unable to install package trimclip, though, which appears not to be part common TeX distributions, and thus I am afraid may constitute a problem down the road, when/if the paper is accepted for publication. So, I've tried to use egreg's ideas to do without a specialized package, and have come up with "half-equal" inequalities which look similar to egreg's, but of course "my solution" is too crude and ad hoc, and it doesn't scale; see the code and results below. Can this "solution" be improved, again without using a rare latex package? Perhaps I should have said where such symbols can be useful -- of course, it's in Fourier analysis of (possibly signed or even complex-valued) measures, where it is natural to define the distribution function of such a measure at every point of its discontinuity to be exactly in the middle between the left and right limits of the distribution function at such a point.

As for the the corresponding "half-equal" delimiters, I haven't been able to get a combination of a half-bracket and half-parenthesis as I would like to, but rather a combination of \lfloor and \rfloor, with which I guess I could live for a while.

Here are "my" code and the corresponding images (sorry, I don't know how to make the images larger):

\documentclass{article}

\usepackage%[dvips]
{color}

%left half of =
\newcommand{\lhalfeq}1{\mbox{$#1=$\kern-3.5pt{\textcolor{white}{\rule[1pt]{4pt}{3.4pt}}} } }

%right half of =
\newcommand{\rhalfeq}1{\mbox{$#1=$\kern-7.9pt{\textcolor{white}{\rule[1pt]{4pt}{3.4pt}}} } }

% greater than or half-equal to
\newcommand{\gtheq}{\mathrel{\mathpalette\xgtheq\relax}}
\newcommand{\xgtheq}[2]{%
\vcenter{\hbox{%
\oalign{$#1>$\cr\lhalfeq{#1}\hidewidth\cr}%
}}
}
% less than or half-equal to
\newcommand{\ltheq}{\mathrel{\mathpalette\xltheq\relax}}
\newcommand{\xltheq}[2]{%
\vcenter{\hbox{%
\oalign{$#1 \rhalfeq{#1}\cr}% }}\kern-3pt } \begin{document}$(-\infty,b\rfloor\kern-3.9pt\lfloor$\qquad %$\rfloor\kern-3.9pt\lfloor a,\infty)$\qquad %$a\gtheq b\ltheq c\$

\end{document}

• Are you presenting another alternative approach your Q ? if not it would be advisable to update your Q with the EDIT/UPDATE: to reach the audience as this space is reserved for only Answers. – texenthusiast Aug 26 '13 at 22:13